Author Topic: Faraday homopolar disc problem  (Read 12739 times)

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Offline Circlotron

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Faraday homopolar disc problem
« on: September 23, 2016, 05:37:30 am »
So we get a C shaped magnet with closely spaced pole pieces and between (but not touching) those pole pieces we put a copper disc with its centre aligned with the magnet pole piece centre. Then we pass a direct current from the centre to the edge of the disc via a slip ring and brush on the edge. If the disc is free, it will rotate around it's centre. Nothing new here. All dates back to the 1830's.

However...

If we grab hold of that disc to prevent it from rotating, we would presume that an equal but opposite torque is exerted on the magnet. If it were not so, we would have set the physics world on its ear with our new reactionless motor.

The only alternative is that that reaction torque sets the magnet rotating. Now the big question is - does this rotating magnet induce a counter EMF in the disc? Present day physics says "no". The trouble is, this would mean that we have produced an electric motor that operates and produces mechanical power when supplied with a current but has no voltage drop across it. (Ignoring resistive losses)

A normal DC motor, when rotating, generates a counter EMF that opposes the supply voltage and therefore allows a predictable current to pass though it, the current being (supply voltage - counter EMF) / motor resistance.

Our homopolar motor with a stationary disc either does NOT produce a counter torque on the magnet structure, or if it does, the rotating magnet does NOT induce a counter EMF in the disc and therefore does not draw any electrical power while producing mechanical work because of zero voltage drop across it.

Neither situation seems satisfactory. Choose your poison.
« Last Edit: September 23, 2016, 08:30:27 am by Circlotron »
 

Offline NiHaoMike

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Re: Faraday homopolar disc problem
« Reply #1 on: September 23, 2016, 05:57:17 am »
Real conductors have resistance that need voltage to overcome. If you try to use a superconductor, it will interact with the magnetic field in such a way as to completely change the situation.

It's a little like the classical "connect a charged capacitor to an uncharged one" problem where calculations show that half the energy disappears. The question now is where the energy goes in an ideal circuit. There would be no resistance to absorb it. One answer is that it gets radiated, but an ideal circuit wouldn't do that either. The real answer is that it's physically impossible to build such an ideal circuit since there will always be inductance. If you could prevent it from radiating, it will just act as a LC tank circuit, never settling to steady state.
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Offline Circlotron

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Re: Faraday homopolar disc problem
« Reply #2 on: September 23, 2016, 08:26:59 am »
Mike, the subject is electromagnetism and whether or not an axially rotating magnetic pole piece induces an EMF across the radius of the disc. I understand what you are saying about resistance and the real world, but it doesn't materially alter the situation I have presented here.
 

Offline John Coloccia

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Re: Faraday homopolar disc problem
« Reply #3 on: September 23, 2016, 10:35:46 am »
Our homopolar motor with a stationary disc either does NOT produce a counter torque on the magnet structure, or if it does, the rotating magnet does NOT induce a counter EMF in the disc and therefore does not draw any electrical power while producing mechanical work because of zero voltage drop across it.

How did you arrive at this conclusion?
 

Offline Circlotron

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Re: Faraday homopolar disc problem
« Reply #4 on: September 23, 2016, 11:09:35 am »
Which part?
 

Online helius

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Re: Faraday homopolar disc problem
« Reply #5 on: September 23, 2016, 11:26:45 am »
I think both parts actually derive from the same premiss, so they are not really a contradiction, but simply begging the question. If you assume that a magnetic field has an axial symmetry, then rotating it around its axis doesn't change anything, so it can neither lead to a back-EMF nor can it be the cause of a reaction torque.
 

Offline amyk

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Re: Faraday homopolar disc problem
« Reply #6 on: September 23, 2016, 11:29:32 am »
 

Offline Circlotron

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Re: Faraday homopolar disc problem
« Reply #7 on: September 23, 2016, 11:34:48 am »
so it can neither lead to a back-EMF nor can it be the cause of a reaction torque.
It 100% most definitely produces a torque on the disc to cause it to try and rotate, as I fact I have seen with my own experiments some years ago. The magnet assembly is effectively the stator of this motor, and to say that this stator does not try and twist the opposite direction to the disc opens up a big can of worms. My precise reason for starting this thread.
 

Offline Circlotron

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Re: Faraday homopolar disc problem
« Reply #8 on: September 23, 2016, 11:42:46 am »
Use the (Lorentz) Force... :D
That level of mathematics is beyond me, but if there is an obvious flaw in the reasoning in my first post then please point it out!
 

Offline alsetalokin4017

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Re: Faraday homopolar disc problem
« Reply #9 on: September 23, 2016, 01:47:39 pm »


How's that for a can of worms....   
« Last Edit: September 23, 2016, 01:49:28 pm by alsetalokin4017 »
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Offline SeanB

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Re: Faraday homopolar disc problem
« Reply #10 on: September 23, 2016, 05:49:48 pm »
Try making a ball bearing motor....
 

Online T3sl4co1l

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Re: Faraday homopolar disc problem
« Reply #11 on: September 23, 2016, 06:06:56 pm »
http://mysite.du.edu/~jcalvert/tech/faraday.htm

;)

Browse the rest of the site BTW -- great education!

The resolution to the riddle is that the magnetic field is a product of rotational symmetry.  Rotating the magnet necessarily, cannot have any effect, because it's already spinning!

Using a horseshoe magnet adds confusion, but the fields remain the same.

Where is the counter-torque generated?  On the loop shorting across the disc: between the axis and the slip ring contact.

It still seems unbelievable, because one can mentally construct a situation where B field is penetrating the disc, and exists nowhere else (or at least, nowhere asymmetric and nonrotating).  But this is a faulty construction!  The laws of electromagnetism do not allow sharp cutoffs to fields, nor is a field allowed to decay over any distance, without also fringing away.

Thus, whether by distance or by shielding, the magnetic field must always be whole (it loops back on itself, because \$\nabla \cdot \vec{B} = 0\$ ).  And anywhere within that field that you might place a return wire, it must necessarily be exposed to the same enclosed flux.  You can equally well ask, what should happen if the disk is nonrotating and the contact is revolved around it?  And again, the answer is the same, a voltage is created.

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Offline John Coloccia

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Re: Faraday homopolar disc problem
« Reply #12 on: September 23, 2016, 06:33:32 pm »
Ah...I understand the question how. The Faraday "paradox". What a crappy way of teaching this or talking about it, to say that there's some paradox when it's really quite simple. When the magnet spins, there's no change in flux. It's a mistake to think of the field as being "attached" to points on the magnet. It's just like if you had a spherical light source and the source was rotating.
 

Offline alsetalokin4017

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Re: Faraday homopolar disc problem
« Reply #13 on: September 23, 2016, 06:58:02 pm »
OK... now hold that thought in your mind while you consider this:

The easiest person to fool is yourself. -- Richard Feynman
 

Offline IanB

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Re: Faraday homopolar disc problem
« Reply #14 on: September 23, 2016, 06:58:28 pm »
But there is a torque on a rotating symmetrical magnet, as seen by my experiment here:

https://youtu.be/0iVOG2eMAGY

There is a force on the wire carrying the current, and since the wire is fixed in space the magnet experiences a reaction force and spins. The magnet poles are oriented up and down, so there is rotational symmetry in this system.
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Online helius

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Re: Faraday homopolar disc problem
« Reply #15 on: September 23, 2016, 08:53:13 pm »
How can you say the torque isn't on the conductive plane (with the painted sectors) between the axis and the perimeter?
 

Online CatalinaWOW

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Re: Faraday homopolar disc problem
« Reply #16 on: September 23, 2016, 10:17:20 pm »
OK... now hold that thought in your mind while you consider this:



Were I to look for an explanation of this I would explore the interaction of the boiling nitrogen with the magnet.  It appears that the magnet is slightly heavy on one side, giving it a mildly preferred orientation wrt gravity.  It is effectively a very low friction pendulum, with a natural frequency of a little over a second.  Which seems to be roughly the bubble rate of the nitrogen.  So now you are driving a high Q mechanical resonator near its resonant frequency.  Minor shape variances on the magnet would cause a preference for one direction over the other. 

With a little money or a well equipped lab you could check this by conductively coupling a colder liquid to the nitrogen bath.  You could slow or eliminate the boiling.  Lots of choices for the colder liquid.  Nitrogen under vacuum.  Neon.  Or go beast mode and use hydrogen.

Or do it like an engineer and use a piece of cardboard to shield the suspended magnet from the gas boiling off and from any spray from the bursting bubbles.
 

Offline alsetalokin4017

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Re: Faraday homopolar disc problem
« Reply #17 on: September 23, 2016, 11:57:36 pm »
I don't have a video of it but the experiments have been done. The magnet rotates even when shielded from the turbulence of the boiling LN2.
The easiest person to fool is yourself. -- Richard Feynman
 

Online T3sl4co1l

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Re: Faraday homopolar disc problem
« Reply #18 on: September 24, 2016, 12:04:42 am »
But there is a torque on a rotating symmetrical magnet, as seen by my experiment here:

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There is a force on the wire carrying the current, and since the wire is fixed in space the magnet experiences a reaction force and spins. The magnet poles are oriented up and down, so there is rotational symmetry in this system.

Obviously, it doesn't matter if:
1. The magnet is in free space, projecting a field through the conductive disk, or
2. the magnet is attached to the disc, above or below or within it, or
3. the magnet IS the disc.

What you've created is only one thing:

A mechanism for the generation of verbal ambiguity.

The torque is acting on the magnet (the whole physical object), because of the magnet (the invisible magnetic field, permeating the physical object and the space around it), and the current flowing through/over the object (the plating and material bulk are conductive).

There is no ambiguity in the physical phenomenon. ;)

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Offline John Coloccia

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Re: Faraday homopolar disc problem
« Reply #19 on: September 24, 2016, 12:15:54 am »
I don't have a video of it but the experiments have been done. The magnet rotates even when shielded from the turbulence of the boiling LN2.

The most obvious conclusion would be that the assumed symmetry of one of the magnets is actually not symmetrical for some reason. It's hard to imagine what else could account for it. The next question would be WHY isn't it symmetrical, and I wouldn't even have a guess. Obviously something is changing, though.

I know strange things happen to the flux lines in superconductors, more than just the Meissner effect depending how it's setup, so maybe it's something to do with the setup?
 

Offline IanB

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Re: Faraday homopolar disc problem
« Reply #20 on: September 24, 2016, 12:37:01 am »
Responding to my video above, both helius and T3sl4co1l have pointed out that current is flowing radially from the edge of the spinning disk magnet to the center, and that there are magnetic field lines running vertically through the magnet between top and bottom faces. The current is therefore intersecting the field lines at right angles and this is generating a force that causes the magnet to spin (the paper disk is just for show). The "mystery" is not a mystery.

It is curious perhaps that the cylindrical magnet as a conductive metal disk, and the cylindrical magnet as a source of magnetic field lines, are two independent components of the system. It is not the "magnet" that spins, but the "conductive metal disk".
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Online T3sl4co1l

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Re: Faraday homopolar disc problem
« Reply #21 on: September 24, 2016, 03:06:24 pm »
So, since the magnet is already spinning (in the sense that the electron spins, within, are producing a coherent magnetic field), riddle this:

If a magnet is suspended in free space, then its temperature raised to the Curie point, say by heating it with random-polarized light), does it begin spinning?

Or in other words, does the magnetization contain (microscopic) angular momentum, which is released as (macroscopic) rotation of the object, when the spins are decoupled (by raising the temperature)?

Tim
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Offline John Coloccia

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Re: Faraday homopolar disc problem
« Reply #22 on: September 24, 2016, 03:33:42 pm »
So, since the magnet is already spinning (in the sense that the electron spins, within, are producing a coherent magnetic field), riddle this:

If a magnet is suspended in free space, then its temperature raised to the Curie point, say by heating it with random-polarized light), does it begin spinning?

Or in other words, does the magnetization contain (microscopic) angular momentum, which is released as (macroscopic) rotation of the object, when the spins are decoupled (by raising the temperature)?

Tim

Angular momentum is conserved. It's either there or it's not, so I'm not sure how it could stored up and then released.
 

Offline amyk

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Re: Faraday homopolar disc problem
« Reply #23 on: September 24, 2016, 04:51:55 pm »
The homopolar generator problem becomes even more fun when relativity is involved. :)
 

Offline TerraHertz

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Re: Faraday homopolar disc problem
« Reply #24 on: September 25, 2016, 01:08:52 am »
I don't have a video of it but the experiments have been done. The magnet rotates even when shielded from the turbulence of the boiling LN2.

Next experiment: suspend the magnet in a vacuum vessel above the superconductor.
My guess is the pendulum torque is due to small differences in surface temp of the magnet acquired from the surrounding temp-gradient gas, and differences in the statistical vector of average force due to gas molecules rebounding from the magnet surface.
Prediction: in a vacuum, it won't rotate.
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